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00:00 - 19:0019:00 - 00:00

7:00 PM
@ACuriousMind I ask you
 
I'm an engineer, I dunno anything
 
Or @Danu
Since you're actually here
 
@barrycarter: I'm around for the next half hour or so if you want to ask anything ...
 
@Slereah Wait what
What rule
 
10 mins ago, by Slereah
Should I do $d^n p \rightarrow -d^n p$ if n is odd
This thing
 
7:03 PM
@JohnRennie Awesome! OK, I want to use the line invariant to compute position for a constantly accelerating reference frame.
 
So all it took was someone who really wanted SRT answer to make you a regular in this chatroom @JohnRennie?
 
I summoned him. It's a Level 4 Spell.
"Summon the Rennie"
 
Line element, or invariant interval ;)
 
The thingy @JohnRennie showed me yesterday.
 
Damn I messed that up :P
 
7:04 PM
I don't expect much from you, @Danu
 
@Danu the chat room has got a lot more interesting in the last few weeks. Last night I spectated at an argument between Barry and 0celo7 about manifolds and actually learned something.
 
@JohnRennie Really?
 
@Slereah It's $-$ for odd, $+$ for even.
I think $\int d^np$ is $+$ $\forall n$.
 
Yes, the formal definition of a chart and it's relevance to coordinates was new to me
 
@JohnRennie ...How.
 
7:06 PM
Sounds about right
 
But you kind of lost me after that
 
How can you know 100,000 rep worth of GR and not know the formal definition of a manifold??
 
@JohnRennie Ah, but probably not useful. OK, if I see an object accelerating constantly, in 'du' time for me, the objects 'dx' is v[u]*du + (an acceleration term). Good so far?
 
@0celo7 you have to understand that my background is as an experimental colloid scientist. I knew absolutely nothing about GR until I joined the Physics SE.
 
@JohnRennie Yeah, but you've never read the classics? MTW, Wald, HE?
 
7:07 PM
And even now my fundamental knowledge of maths is appallingly bad
Have you ever tried reading MTW?
 
Are you interested in changing that, or not really?
 
Don't iconoclastify yourself!
 
@JohnRennie A section here and there.
@JohnRennie We can help you.
 
Me too. For example I read chapter 6 to learn about accelerated motion in SR - which his Barryness is about to benefit from :-)
But as a whole it's utterly indigestible
 
I dub thee Sir @JohnRennie
 
7:08 PM
@JohnRennie Let me see.
Let's get out the legal copy
 
Carroll is the only real GR book :3
 
@Danu Leave.
 
@Danu to be honest I like playing in the shallow waters of GR
 
eye twitch
@JohnRennie What part in MTW is confusing you
 
It would be nice to be able to swin down into the deep water, but I doubt I have the concebntration to learn that much maths these days
 
7:10 PM
@JohnRennie Carroll :3
 
Don't listen to him @JohnRennie
Read Sachs and Wu
 
Bad notation, bad remarks about the nature of mathematics and physics ;)
 
Phew, momentum proven
 
Physicists seem to place great stock in books.
@JohnRennie Since I get distracted easily, I won't fault you, but let me know when you're ready.
 
@Danu Also they require you to be familiar with Riemannian geometry before reading it
3
A: Getting started self-studying general relativity

0celo7This list is extensive, but not exhaustive. Special Relativity E. Gourgoulhon (2013), Special Relativity in General Frames. This is a rigorous and encyclopedic treatment of special relativity. It contains pretty much everything you will ever need in special relativity, like the Lorentz fac...

 
7:11 PM
@0celo7 Take chapter 6: they could have said "we're looking for a metric where the norm of the four-acceleration is $a$ and here's how we do it"
 
@JohnRennie I have a nice list
Right there
For you to look at
 
Instead they assume that's obvious - but it isn't
 
@JohnRennie What page
 
Anyway @barrycarter what did you want to know?
 
Because that doesn't make much sense to me
What four-velocity
And why would they call that $a$
Very confusing
 
7:12 PM
Oops, I meant four-acceleration
 
...of what?
 
@JohnRennie Oh, if an object has constant acceleration in its own frame, my observation of its motion in du time is v[u]*du + (some acceleration term), right?
 
of the observer at the origin
@barrycarter it's a surprisingly involved calculation
 
@JohnRennie why don't you want to learn some math :(
 
@JohnRennie Well, I see the acceleration term as a t dt but notice I actually mean t not u. The acceleration is constant inthe other frame.
 
7:14 PM
@barrycarter Objects do not accelerate in their own frame
 
@Danu You know what I mean :P
 
No, I don't really...
 
One way is to note that in the Earth frame the acceleration of a moving observer is a/gamma^3, where a is the acceleration measured by the accelerating observer
 
According to MTW that's not true @Danu
 
@Danu Meaning that the change in velocity for a third party observer is based on the object's frame.
 
7:16 PM
hi
 
@0celo7 Is there some mix-up of definitions going on? I would say the obvious way to define the frame "of an object" is the one where the object is always at the origin.
 
I think the rest frame they're talking about is where the four-velocity is the timelike basis vector
 
I'm pretty sure everybody agrees on that definition.
 
We're going to get nothing done with all this kibbitzing!
 
And since it's me that's learning, that's a bad thing!
 
7:17 PM
@Danu Yes, but the acceleration cannot be zero in that frame if it's not zero in other frames
Because it's tensorial
 
@JohnRennie You can just ignore us
@0celo7 Lol.
If an object is always at the origin, what is its acceleration?
 
I'm saying that, from my point of view, the dx of the object depends partly on t and dt in the other frame. Is that correct?
 
@Danu it's the acceleration with which an object I release accelerates away from me
 
@Danu We're talking infinitesimals here, remember?
 
Like right now my acceleration is 9.81 m/s^2
 
7:18 PM
@JohnRennie I like that. Since your own frame isn't inertial, that would sort of work.
 
@JohnRennie If you're releasing something, that's something different. I was just objecting against something "accelerating in its own rest frame"
@JohnRennie In your frame it's zero.
 
FFS we have language police as well as maths purists
 
@Danu As opposed to constant acceleration in the observer's frame, which leads to a paradox of sorts.
 
@JohnRennie In any case, I think the take-home message is: Ignore side-tracks in the conversation ;)
 
@Danu Is the velocity vector $u=0$ in the rest frame?
 
7:19 PM
I probably should have done some background. I'm saying that: v du + a u du is NOT the correct dx for the remote observer.
 
@barrycarter: you can select any convenient system of coordinates and use it to calculate the four accelaration of an object e.g. me.
 
@0celo7 I didn't follow the conservation---I don't know what "the velocity vector" is, nor what "the rest frame" is.
 
The norm of this four-acceleration is the acceleration the observer feels in their rest frame
 
I also have no interest in discussing this any further---I don't think there is anything to learn here.
 
@Danu You were literally just talking about the rest frame
 
7:20 PM
i.e. Never mind people!
 
@Danu What the fuck?
 
@JohnRennie OK, I may've jumped ahead. In the object's own frame, it's position is (t,0,0,0), so its norm-like thing is -t^2, right?
 
@0celo7 I never said "rest frame", IIRC.
 
@Danu "its frame" then
 
The problem is that the metric for the rest frame of an accelerating observer is not the Minkowski metric, it is the Rindler metric.
 
7:22 PM
@JohnRennie So I can't say the object's norm is -dt^2 in its own frame?
 
I'm abusing the terminology a bit because they are the same metric, but written in different coordinates
 
I thought ds^2 worked regardless of frame acceleration?
 
Have a look at the calculation of four acceleration for an observer in the Schwarzschild metric done here:
17
Q: What is the weight equation through general relativity?

user17093The gravitational force on your body, called your weight, pushes you down onto the floor. $$W=mg$$ So, what is the weight equation through general relativity?

 
@JohnRennie That appears to be fairly long, so let me just ask: are you saying that in my setup, ds^2 is not constant across observers?
 
If the metric is g_\mu\nu then the norm of the four acceleration a^\mu is g_\mu\nu a^\mu a^\nu
@barrycarter ds^2 is still invariant, but you calculate ds^2 using the metric and the Earth and accelerating observer use different metrics
 
7:25 PM
@JohnRennie Er, there's no gravity involved in my problem. Just a constant acceleration object.
 
Errr wait
 
@barrycarter hasn't anyone explained the equivalence principle :-)
 
@JohnRennie You're the only one who has guided me :P
 
@Slereah what have I done wrong now?
 
The fancy normalization they give for momentum states is $\langle p \vert q \rangle = 2 E_p (2\pi)^3 \delta(p-q)$
 
7:27 PM
@JohnRennie You always do this, and I can't resist... Invariant!
 
God it's like being at school again!
 
(don't kill me)
 
But is that normalized?
The integral over $p$ will just be $2E_q$
 
@Danu damn, and it's too late to edit now :-)
 
@JohnRennie Sometimes, miracles happen ;)
 
7:28 PM
Shouldn't it normalize to 1
I know they are not wavefunctions stricto sensu, but still
 
@Danu how did you do that?
 
@JohnRennie I thought the entire point of using ds^2 (instead of s^2 as some other treatments do) is to allow for acceleration?
 
@JohnRennie I'm a moderator over at History of Science and Mathematics, which comes with network-wide moderation powers in chat...
 
@Danu Cool :-)
@barrycarter: if you want to understand relativity you need to go off and read chapter 6 of Misner, Thorne and Wheeler. I can help you work through it, but it's involved enough that I can't do it justice in this chat room.
 
@JohnRennie OK, bummer. I thought I had this down.
 
7:31 PM
Basically you can calculate the acceleration experienced by an observer from the metric.
 
Well, that's what I'm trying to do.
 
So you have to look for a metric in which the observer at the origin has a constant acceleration
 
It just gets complicated because the observer's metric must use the object's coordinates.
 
And this turns out to be the Rindler metric
 
I still think it might be do-able from the invariance of the Minkowsky metric.
You just end up solving a differential equation.
 
7:33 PM
I haven't seen it done that way, but yes I'm sure it is possible
 
OK, that's what I was going for. I wanted your help before I setup the equation.
 
I think you will save us both much pain if you read MTW first. Also check out the Physics FAQ article:
for a summary of the results.
 
@JohnRennie I've read that (page, not book) many many times. I'm just trying to derive those results from first principles. I've sort of done it in some questions, but I'm looking for a clean non piecemeal derivation.
 
Maybe even check out these superbly written and easy to understand answers:
 
LOL :)
 
7:38 PM
@barrycarter MTW show how the equations are derived.
 
I want to try out my dfq solution first. If it works, I want to continue using the ds^2 thingy.
@JohnRennie Yes, but I'm not big into books.
And I'm ok with accepting the invariance of ds^2 as a postulate.
 
Wait how do I even prove that $\langle 0 \vert a_p a^\dagger_q \vert 0 \rangle = (2\pi)^3 \delta(p-q)$
Hm
I guess since $a_p\vert 0 \rangle = 0$ I can use the commutator
 
I guess it's fun (and instructional) to do these things for yourself, but in this case I suspect the pain will outweigh the gain
 
@barrycarter you can prove that
It's in Zee, probably in Carroll too
 
@barrycarter the trouble is you need the metric to compute ds^2, and you don't know the metric in the accelerating coordinates
 
7:40 PM
Landau
 
Ocelot, yes, I realize you CAN prove it, but I'm saying I don't need to. I want to derive from that starting point.
 
yeah nvm I got it
 
@JohnRennie For the object itself, it's always just -dt^2, right?
I have tons of time, and I love math. So, I'll try it until I get sick of it.
 
In an observer's rest frame their ds^2 = -dt^2 yes
 
@JohnRennie The trick is to find ds^2 in someone else's frame for the original observer/object.
 
7:43 PM
@barrycarter and that's what you need the metric for
 
@JohnRennie And you can't just use the obvious one.. but do you?
@JohnRennie Isn't it v[u] du + a t dt for dx?
Where v[u] is the velocity at time u
 
Oh
Apparently the measure too has to be Lorentz invariant
nvm for also the previous question
 
Yes, but if you're doing the calculation in the Earth frame you need a to be the acceleration in the Earth coordinates, and that's not the same as the acceleration in the accelerating observer's coordinates
 
Much like Caillou everyday I'm learning something new
 
@JohnRennie By earth frame, you mean an inertial frame, right?
 
7:45 PM
@barrycarter Yes aka the non-accelerating frame
 
@JohnRennie Correct, but notice I'm using a t dt, and NOT du. In other words, I AM using the object's coordinates.
 
You've lost me know, and in any case I need to get to bed. 6 a.m. start tomorrow (again).
 
OK, have fun :) I'll run the math and see what happens
 
Actually I think having to learn GR without a rigorous grounding in the maths forces you into understanding the physics i.e. what's actually going on. I suspect that's actually harder than working through the maths - it's certainly a prolonged and painful process.

However once you've gained an understanding, at least a partial understanding, of the physics it puts you in a superb position to answer questions from GR novices in an intuitively obvious way. I haven't check exatly what fraction of my rep comes from the GR tag, but I bet if you looked closely it was almost all due to simple ans
 
Woo
2.3 finished
Onto 2.4
Only 700 pages left
 
8:06 PM
@Slereah I'm going to begin reading peskin and schroeder in a month so can you leave behind some stuff for me to look at kind of like what you're doing now?
like things that are unclear.
 
Well ask me I guess?
It's a common thing in textbooks to not have complete proofs
 
@JohnRennie Ultimately, I believe the universe is running a mathematical equation. Using that equation, we can predict what will happen and what we will observe. However, there is no ultimate meaning behind it all. IE, physics is really nothing more than math.
 
@Slereah yea, and thanks I will.
 
I need to solve $x^{n-1}=1/\epsilon$ for $n$ and I don't know how.
:(
 
Between cleaning, working out and QFT this is a Productive Week End
Use the log
 
8:16 PM
Which one
 
The log arithm
 
Do I need
Abramowitz and Stegun
 
Yes, the logarithm is some PhD level math
 
@Slereah wait a moment
wait JUST one moment
 
I have awaited a moment
I can't wait any longer, sorry
 
8:27 PM
wait
is $\ln x^y=y\ln x$ correct
 
yes
 
crazy shit dude
 
Did you just discover the logarithm
 
yes.
@Slereah wow logs are amazing
How come we never learn about them
 
I know right
Well you do, if you go to school
Man I learned about logs in high school m8
That's why the US has terrible math scores
 
8:31 PM
wtf
I've never heard of a log
4
reading the wiki article right now
so interesting
 
With @0celo7 I can't tell if he's trolling me
He has odd gaps in his knowledge
 
have I ever trolled?
 
Yes.
 
wtf bro
@Slereah like what
 
You have been mean to Duffield
v. inappropriate
 
8:34 PM
what does Duffield have to do with anything
@yuggib These analysis proofs are fun
 
8:53 PM
Even engineers know about logs, so I think Ocelot is trolling.
 
not all of us
 
Slide rules are based on logs.
 
@barrycarter Maybe
 
I'm pretty sure they are.
 
I love figuring out analysis problems as I'm complaining about how it's impossible to prove
 
9:06 PM
@0celo7 5 in ap calc without a log, yeah right.
unless you forgot.
D:
 
No one forgets his first logarithm
 
9:19 PM
@3075 I actually did
 
In freshman physics we consider a 1D infinite square well.
We usually consider the case where we have "one particle".
We learn that this particle can be one of various states.
For example, we suppose the particle can be in state 0, 1, 2, ... etc.
If we have multiple electrons, we say the levels get filled up from the bottom upward, with one electron per level.
What happens if we fill the thing with bosons?
 
9:38 PM
what are you talking about
you did QM in freshman physics?
 
9:53 PM
I did some QM in the last year of high school
It was pretty lowkey, though
Just like "hydrogen has discrete levels, compute some energy for transitions"
 
guys, how goes it
 
 
1 hour later…
11:13 PM
Hello all…
Has anyone here read Quantum Mechanics: The Theoretical Minimum?
I'm beginning to suspect that understanding the mathematical concepts described within may require a solid understanding of linear algebra…
…and I'm hoping to confirm
Went back and skimmed intro…book requires basic knowledge of linear algebra. Unfortunately when I originally bought the book I thought that linear algebra = high-school algebra.
 
11:43 PM
/me heads off to purchase linear algebra for dummies
 
@MonaLisaOverdrive linear algebra is like the god of math for physicists
get very good with linear algebra
 
I was afraid you'd say that.
 
@MonaLisaOverdrive get VERY good
you need not be afraid youngling
 
I find your use of the term 'youngling' amusing
 
linear algebra kinda is really 5 or 6 concepts repeated again in a million ways to make you understand how universal it is
 
11:47 PM
So…if I get a good understanding of the 5 or 6 concepts, I can re-tackle The Theoretical Minimum?
'Cause my time is rather…restricted
 
im not probably really doing it justice, but basically what you do is identify what the concept of a vector space is, its a first real dive into the structure of math
and getting good with understanding linear algebra will save you time all the way through the rest of your physics learning
so give it some real attention, since basically most of the things you will do with physics have analogs in linear algebra notation,
Hey @ACuriousMind is there anything this article is actually getting wrong about the holographic universe models motherboard.vice.com/read/…
 
Welp…back to the drawing board.
 
@MonaLisaOverdrive you know how currently energy conservation seems like something really important to you
 
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